The Laws of Thermodinamics

1. The Laws of Thermodinamics

2. Work in Thermodynamic process • Energy can be transferred by heat and by work done on the system • W=-FΔy =-PA Δy • The work W done on a gas at constant pressure is given by: W= -P ΔV P-pressure through the gas ΔV- the change in volume

3. W= -P ΔV –can be used to calculate the work done on the system only when the pressure of the gas remain ct. during the expansion or compression ISOBARIC PROCESS- a process in which the pressure remain constant • The area under the graph in a PV diagram is equal in magnitude to do work done on a gas: A=PΔV

4. The first law of thermodynamics (energy conservation law that relates changes in internal energy): If a system undergoes a change from an initial state to a final state, where Q is the energy transferred to the system by heat and W is the work done on the system, the change in the internal energy of the system,ΔU is given by: ΔU= Uf-Ui = Q+W

5. Q>0- when energy is transferred into the system by heat • Q<0- when energy is transferred out of the system by heat • W>0 –when work is done on the system • W<0 – when the system does work on its environment • The internal system of any isolated system must remain constant, ΔU=0 • If the system isn’t isolated, ΔU=0if the system goes through a cyclic process, which P, V, T, n (nr of moles) return to original values

6. U=3/2 n R T • ΔU=3/2 n R ΔT • Molar specific heat at V=ct. of a monatomic ideal gas, Cv is: Cv =3/2 R • ΔU=n CvΔT • Q= ΔU-W= ΔU+ P ΔV • P ΔV=n R ΔT • Q= 3/2n R ΔT+ n R ΔT= 5/2 n R ΔT • Molar heat capacity at P=ct., Cp is: Cp=5/2R • Q=n CpΔT • Cp = Cv +R

8. Adiabatic process- no energy enters or leaves the system by heat (can still do work) • Q=0 ΔU=W • For an ideal gas undergoing an adiabatic process: PVγ =ct γ= Cp / Cv γ - adiabatic index of the gas

9. Isothermal process- the temperature of the system doesn’t change • ΔT=0; ΔU=0; W=-Q • For an ideal gas at T=ct P=nRT/V • The work done on the environment during an isothermal process is given by: Wenv=nRTln(Vf/Vi) • The PV diagram of a typical isothermal process , contrasted with an adiabatic process

12. A heat engine takes energy by heat and converts into electrical or mechanical energy • A heat engine carries some working substance through cyclic process during which: • 1. energy is transferred by heat from a source at a high temperature • 2. work is done by the engine • 3. energy is expelled by the engine by heat to a source at lower temperature

13. The absorbs energy Qh from a hot reservoir, does work Weng , that gives up energy Qc to the cold reservoir • W= -Weng (work is done on the engine) • ΔU=0=Q+W Q net=-W =Weng • The work done by the engine= the net energy absorbed by the engine • Q net=|Qh|-|Qc| • The thermal efficiencye of a heat engine is the work done by the engine divided by the energy absorbed during one cycle: • e= Weng / |Qh|=|Qh|-|Qc|/ |Qh|=1-|Qc|/|Qh| • e=1- engine has a 100% efficiency

14. The heat pumps and refrigerator The coefficient of performance for a refrigerator or an air conditioner (the cooling mode): COP=|Qc|/W • The coefficient of performance of a heat pump operating in a heat mode : COP=|Qh|/W

15. THE SECON LAW OF THERMODINAMICS: No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for the performance of na equal amount of work (e<1) Ist law: we can’t get a greater amount of energy out of a cyclic process that we put in 2nd law: we can’t break it even

16. Reversible process- every state along the path in an equilibrium state, so the system can return to its initial conditions by going along the same path in the reverse direction. • Irreversible process- a process that doesn’t satisfy this requirement • Most natural processes are known to be ireversible

17. The Carnot Engine: a heat engine operating an ideal, reversible cycle (the Carnot cycle) between two reservoirs is the most efficient engine possible • Carnot’s theorem: no real engine operating between 2 energy reservoirs can be more efficient than a Carnot engine operating between the same 2 reservoirs

18. Carnot Cycles:

19. An ideal gas is contained in a cylinder with a movable piston at one end • 1. The process A→B is an isothermal expansion at temperature T (the gas is placed in thermal contact with a hot reservoir, and the gas absorbs energy Qh from the reservoir and does work WAB in raising the piston) • 2. The process B→C , the gas expends adiabatically (energy enter or leave by heat). The temperature falls from Th to Tc, and the gas does work WBC to the raising piston

20. 3. The process C→D, the gas is placed in thermal contact with a cold reservoir at temperature Tc and is compress isothermally at temperatures Tc (the gas expels energy Qc to the reservoir and the work done on the gas is WCD) • 4. The process D→A , the gas is compressed adiabatically (the temperature increases to Th, and the work done on the gas is WDA)

21. |Qc|/|Qh|=Tc/Th • e=1-Tc/Th • All carnot engines operating reversibly between the same 2 temperatures have the same efficiency The third Law of Thermodynamics: its impossible to lower the temperature of a system to absolute zero (such reservoirs are not available) All real engines operate irreversibly, due to friction and the brevity of their cycles, are less efficient than the Carnot engine

22. Entropy • Let Qr be the energy absorbed or expelled during a reversible, constant temperature process between two equilibrium states. Then the change in entropy during any constant temperature process connecting the two equilibrium states is defined as: ΔS =Qr/T SI unit: J/K (Joules/Kelvin)